Problem: The lifespans of meerkats in a particular zoo are normally distributed. The average meerkat lives $13.1$ years; the standard deviation is $1.5$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a meerkat living less than $14.6$ years.
$13.1$ $11.6$ $14.6$ $10.1$ $16.1$ $8.6$ $17.6$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $13.1$ years. We know the standard deviation is $1.5$ years, so one standard deviation below the mean is $11.6$ years and one standard deviation above the mean is $14.6$ years. Two standard deviations below the mean is $10.1$ years and two standard deviations above the mean is $16.1$ years. Three standard deviations below the mean is $8.6$ years and three standard deviations above the mean is $17.6$ years. We are interested in the probability of a meerkat living less than $14.6$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the meerkats will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the meerkats will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $11.6$ years and the other half $({16\%})$ will live longer than $14.6$ years. The probability of a particular meerkat living less than $14.6$ years is ${68\%} + {16\%}$, or $84\%$.